# Find the angle between the lines

Question:

Find the angle between the lines $\mathrm{y}=(2-\sqrt{3})(\mathrm{x}+5)$ and $\mathrm{y}=(2+\sqrt{3})(\mathrm{x}-7)$

Solution:

Given equations are $y=(2-\sqrt{3})(x+5)$ and $(2+\sqrt{3})(x-7)$ The given equation can be written as

$\Rightarrow y=(2-\sqrt{3}) x+(2-\sqrt{3}) 5 \ldots \ldots 1$

$\Rightarrow y=(2+\sqrt{3}) x-7(2+\sqrt{3}) \ldots \ldots .2$

Now, we have to find the slope of equation 1

Since, the equation 1 is in $y=m x+b$ form, we can easily see that the slope

$\left(m_{1}\right)$ is $(2-\sqrt{3})$

Now, the slope $\left(\mathrm{m}_{2}\right)$ of equation 2 is $(2+\sqrt{3})$ Let $\theta$ be the angle between the given two lines.

$\tan \theta=\left|\frac{m_{1}-m_{1}}{1+m_{1} m_{2}}\right|$

Putting the values of $m_{1}$ and $m_{2}$ in above equation, we get

$\Rightarrow \tan \theta=\left|\frac{2-\sqrt{3}-(2+\sqrt{3})}{1+(2-\sqrt{3})(2+\sqrt{3})}\right|$

$\Rightarrow \tan \theta=\left|\frac{2-\sqrt{3}-2-\sqrt{3}}{1+\left[(2)^{2}-(\sqrt{3})^{2}\right]}\right|$

$\left[\because(a-b)(a+b)=\left(a^{2}-b^{2}\right)\right]$

$\Rightarrow \tan \theta=\left|\frac{2-\sqrt{3}-(2+\sqrt{3})}{1+(2-\sqrt{3})(2+\sqrt{3})}\right|$

$\Rightarrow \tan \theta=\left|\frac{2-\sqrt{3}-2-\sqrt{3}}{1+\left[(2)^{2}-(\sqrt{3})^{2}\right]}\right|$

$\left[\because(a-b)(a+b)=\left(a^{2}-b^{2}\right)\right]$

$\Rightarrow \tan \theta=\left|\frac{-2 \sqrt{3}}{1+[4-3]}\right|$

On simplifying we get

$\Rightarrow \tan \theta=\left|\frac{-2 \sqrt{3}}{1+1}\right|$

$\Rightarrow \tan \theta=\left|\frac{-2 \sqrt{3}}{2}\right|$

$\Rightarrow \tan \theta=\sqrt{3}$ or $-\sqrt{3}$

$\Rightarrow \theta=\tan ^{-1}(\sqrt{3})$ or $(-\sqrt{3})$

$\Rightarrow \theta=60^{\circ}$ or $120^{\circ}$

Hence, the required angle is $60^{\circ}$ or $120^{\circ}$.